Chapter 7 Rational and Irrational Numbers

Chater 7 Rational and Irrational Numbers In this chater we first review the real line model for numbers, as discussed in Chater 2 of seventh grade, by recalling how the integers and then the rational numbers are associated to oints in the line. Having associated a oint on the real line to every rational number, we ask the question, do all oints corresond to a rational number? Recall that a oint on the line is identified with the length of the line segment from the origin to that oint (which is negative if the oint is to the left of the origin). Through constructions (given by tilted squares), we make an observation first made by the Pythagorean society 2500 years ago that there are lengths (such as the diagonal of a square with side length 1) that do not corresond to a rational number. The construction roduces numbers whose squares are integers; leading us to introduce the symbol A to reresent a number whose square is A. We also introduce the cube root 3 V to reresent the side length of a cube whose volume is V. The technique of tilted squares rovides an oortunity to observe the Pythagorean theorem: a 2 + b 2 = c 2, where a and b are the lengths of the legs of a right triangle, and c is the length of the hyotenuse. In the next section we return to the construction of a square of area 2, and show that its side length ( 2) cannot be equal to a fraction, so its length is not a rational number. We call such a number an irrational number. The same argument works for 5 and other lengths constructed by tilted squares. It is a fact that if N is a whole number, either it is a erfect square (the square of an integer), or N is not a quotient of integers; that is N is an irrational number. In the next section we turn to the question: can we reresent lengths that are not quotients of integers, somehow, by numbers? The ancient Greeks were not able to do this, due mostly to the lack of an aroriate system of exressing lengths by their numerical measure. For us today, this e ective system is that of the decimal reresentation of numbers (reviewed in Chater 1 of seventh grade). We recall from grade 7 that a rational number is reresented by a terminating decimal only if the denominator is a roduct of twos and fives. Thus many rational numbers (like 1/3,1/7, 1/12,...) are not reresented by terminating decimals, but they are reresented by reeating decimals, and similarly, reeating decimals reresent rational numbers. We now view the decimal exansion of a number as roviding an algorithm for getting as close as we lease to its reresenting oint on the line through reeated subdivisions by tenths. In fact, every decimal exansion reresents a oint on the line, and thus a number, and unless the decimal exansion is terminating or reeating, it is irrational. The question now becomes: can we reresent all lengths by decimal exansions? We start with square roots, and illustrate Newton s method for aroximating square roots: Start with some reasonable estimate, and follow with the recursion a new = 1 2 a old + N!. a old Through examles, we see that this method roduces the decimal exansion of the square root of N to any required 8MF7-1 2014 University of Utah Middle School Math Project in artnershi with the

degree of accuracy. Finally, we oint out that to do arithmetic oerations with irrational as well as rational numbers, we have to be careful: to get within a secified number of decimal oints of accuracy we may need much better accuracy for the original numbers. Section 7.1: Reresenting Numbers Geometrically First, let us recall how to reresent the rational number system by oints on a line. With a straight edge, draw a horizontal line. Given any two oints a and b on the line, we say that a < b if a is to the left of b. The iece of the line between a and b is called the interval between a and b. It is imortant to notice that for two di erent oints a and b we must have either a < b or b < a. Also, recall that if a < b we may also write this as b > a. Pick a oint on a horizontal line, mark it and call it the origin, denoted by 0. Now lace a ruler with its left end at 0. Pick another oint (this may be the 1 cm or 1 in oint on the ruler) to the right of 0 and denote it as 1. We also say that the length of the interval between 0 and 1 is one (er one unit). Mark the same distance to the right of 1, and designate that endoint as 2. Continuing on in this way we san associate to each ositive integer a oint on the line. Now mark o a succession of equally saced oints on the line that lie to the left of 0 and denote them consecutively as 1, 2, 3,... In this way we can imagine all integers laced on the line. We can associate a half integer to the midoint of any interval; so that the midoint of the interval between 3 and 4 is 3.5, and the midoint of the interval between 7 and 6 is 6.5. If we divide the unit interval into three equal arts, then the first art is a length corresonding to 1/3, the first and second arts corresond to 2/3, and indeed, for any integer, by utting coies end to and on the real line (on the right of the origin is > 0, and on the left if < 0), we get to the length reresenting /3. We can relace 3 by any ositive integer q, by constructing a length which is one qth of the unit interval. In this way we can identify every rational number /q with a oint on the horizontal line, to the left of the origin if /q is negative, and to the right if ositive. The number line rovides a concrete way to visualize the decimal exansion of a number. Given, say, a ositive number a, there is an integer N such that N ale a < N + 1. This N is called the integral art of a. If N = a, we are done. If not, divide the interval between N and N + 1 into ten equal arts, and let d 1 be the number of arts that fit in the interval between N and a. This d 1 is a digit (an integer between (and ossibly one of) 0 and 9). This is the tenths art of a, and is written N.d 1. If N.d 1 = a, we are through. If not reeat the rocess: divide the interval between N.d 1 and N.(d 1 + 1) into ten equal arts, and let d 2 be the number of these arts that fit between N.d 1 and a. This is the hundredths art of a, denoted N.d 1 d 2. Continuing in this way, we discover an increasing sequence of numerical exressions of the form N.d 1 d 2 that get closer and closer to a, thus roviding an e ective rocedure for aroximating the number a. As we have seen in grade 7, this rocess may never (meaning in a finite number of stes ) reach a, as is the case for a = 1/3, 1/7 and so on. Now, instead of looking at this as a rocedure to associate a decimal to a number, look at it as a rocedure to associate a decimal to a length on the number line. Now let a be any oint on the number line (say, a ositive oint). The same rocess associates a decimal exansion to a, meaning an e ective way of aroximating the length of the interval from 0 to a by a number of tenths of tenths of tenths (and so on) of the unit interval. We begin this discussion by examining lengths that can be geometrically constructed; leading us to an answer to the question: are there lengths that cannot be reresented by a rational number? To do this, we need to move from the numerical reresentation of the line to the numerical reresentation of the lane. Using the number line created above, draw a erendicular (vertical) line through the origin and use the same rocedure as above with the same unit interval. Now, to every air of rational numbers (a, b) we can associate a oint in the lane: Go along the horizontal (the x-axis) to the oint a. Next, go a (directed) length b along the vertical line through a. This is the oint (a, b). 2014 University of Utah Middle School Math Project in artnershi with the 8MF7-2

Examle 1. In Figure 1 the unit lengths are half an inch each. To the nearest tenth (hundredth) of an inch, aroximate the lengths AB, AC, BC. 8 7 6 5 4 3 2 1 A C 0 0 1 2 3 4 5 6 7 8 Figure 1 B Solution. We use a ruler to aroximate the lengths. There will be a scale issue: the length of a side of a box may not measure half an inch on our ruler. After calculating the change of scale, one should find the length of AC to be about 3.05 in. Now, it is imortant to know that, by using a ruler we can always estimate the length of a line segment by a fraction (a rational number), and the accuracy of the estimate deends uon the detail of our ruler. The question we now want to raise is this: can any length be described or named by a rational number? The coordinate system on the lane rovides us with the ability to assign lengths to line segments. Let us review this through a few examles. Examle 2. In Figure 2 we have drawn a tilted square (dashed sides) within a horizontal square. If each of the small squares bounded in a solid line is a unit square (the side length is one unit), then the area of the entire figure is 2 2 = 4 square units. The dashed, tilted square is comosed of recisely half (in area) of each of the unit squares, since each of the triangles outside the tilted square corresonds to a triangle inside the tilted square. Thus the tilted square has area 2 square units. Since the area of a square is the square of the length of a side, the length of each dashed line is a number whose square is 2, denoted 2. Figure 2 8MF7-3 2014 University of Utah Middle School Math Project in artnershi with the

We will use this symbol A (square root) to indicate a number a whose square is A: a 2 = A. Since the square of any nonzero number is ositive (and 0 = 0), A makes sense only if A is not negative. Since 2 2 = 4, 3 2 = 9, 4 2 = 16, 5 2 = 25, the integers 4, 9, 16, 25 have integers as square roots. A ositive integer whose square root s a ositive integer is called a erfect square. For other numbers (such as 2, 3, 5, 6,...), we still have to find a way to calculate the square roots. This strategy, of tilted squares, gives a way of constructing lengths corresonding to square roots of all whole numbers, as we shall now illustrate. Examle 3. Figure 3 In Figure 3 the large square has side length 3 units, and thus area of 9 square units. Each of the triangles outside the tilted square is a 1 2 right triangle, so is of area 1. Thus the area of the tilted square is 9 4 = 5, and the length of the sides of the tilted square is 5. Figure 4 By the same reasoning: the large square in Figure 4 is 7 7 so has area 49 square units. Each triangle outside the tilted square is a right triangle of leg lengths 3 and 4, so has area 6 square units. Since there are four of these triangles, this accounts for 24 square units, and thus the area of the tilted triangle is 49 24 = 25 square units. Since 25 = 5 2, the side of the tilted square has length 5 units. That is, 25 = 5. Extension. Here we are using the tilted square construction in order to realize square roots of whole numbers as secific lengths, with the end goal of finding a way to reresent those lengths numerically. (That is the long way of saying that eventually we will find a rocedure to generate the decimal exansion of square roots). Some students may observe that the construction can be generalized to any initially given lengths a and b. They are right, and the generalization is the Pythagorean Theorem that relates the lengths of the sides of a right triangle. In Chater 8, we will make this observation, and elaborate further on the Pythagorean theorem. The following examle rovides a review of that discussion. 2014 University of Utah Middle School Math Project in artnershi with the 8MF7-4

a II b c c a I b II IV a I a b III IV a b III a b b Figure 5 Figure 6 Examle 4. For any two lengths a and b, draw a square of side length a + b as shown in Figure 5. Now draw the dashed lines as shown in that figure. The figure bounded by the dashed lines is a square; denote its side length by c. Then the area of the square bounded by the dashed lines is c 2. Each of the triangles in Figure 5 is a right triangle of leg lengths a and b and hyotenuse length c. Now, move to Figure 6 which subdivides the a + b-sided square in a di erent configuration: the bottom left corner is filled with a square of side length b, and the uer right corner, by a square of side length a. The rest of the big square of Figure 6 is a air of congruent rectangles. By drawing in the diagonals of those rectangles as shown, we see that this divides the two rectangles into four triangles, all congruent to the four triangles of Figure 5. Thus what lies outside these triangles in both figures has the same area. But what remains in Figure 5 is a square of area c 2, and what remains in Figure 6 are the two squares of areas a 2 and b 2. This result is: The Pythagorean Theorem: a 2 + b 2 = c 2 for a right triangle whose leg lengths are a and b and whose hyotenuse is of length c. In this theorem we find the lengths of the sides of tilted squares algebraically. For examle, the tilted square in Figure 1 has side length c where c is the length of the hyotenuse of a right triangle whose leg lengths are both 1: c 2 = 1 2 + 1 2 = 2, so c = 2. Similarly for Figure 2: c 2 = 1 2 + 2 2 = 5, so c = 5. For Figure 3, we calculate c 2 = 3 2 + 4 2 = 9 + 16 = 25, so c = 25 = 5. End Extension 8MF7-5 2014 University of Utah Middle School Math Project in artnershi with the

Section 7.2: Solutions to Equations Using Square and Cube Roots Use square root and cube root symbols to reresent solutions to equations of the form x 2 = and x 3 =, where is a ositive rational number. Evaluate square roots of small erfect squares and cube roots of small erfect cubes. 8.EE.2, first art. In the receding section we introduced the symbol A to designate the length of a side of a square of area A. Similarly, the side length of a cube of volume V as 3 V (called the cube root of V). Another way of saying this is that A is the solution of the equation x 2 = A and 3 V is the solution of the equation x 3 = V. We also defined a erfect square as an integer whose square root is an integer. Similarly an integer whose cube root is an integer is a erfect cube. Here is a table of the first few erfect squares and cubes. Number 1 2 3 4 5 6 7 8 9 Square 1 4 9 16 25 36 49 64 81 Cube 1 8 27 64 125 216 343 512 729 For numbers that are not erfect squares, we can sometimes use factorization to exress the square root more simly. Examle 5. a. Since 100 is the square of ten, we can write 100 = 10. But we could also first factor 100 as the roduct of two erfect squares: 100 = 4 25 = 4 25 = 2 5 = 10. b. Similarly, 729 = 9 81 = 9 81 = 3 9 = 27. Also, 729 = 27 2, and 3 3 = 27, so 3 3 3 3 729 = 27 27 = 27 27 = 3 3 = 9. c. Of course not every number is a erfect square, but we still may be able to simlify: 72 = 36 2 = 36 2 = 6 2. d. 32 = 4 4 2 = 4 2. Similarly, we can try to simlify arithmetic oerations with square roots: e. 6 12 = 6 6 2 = 6 6 2 = 6 2. f. 2 + 8 = 2 + 4 2 = 2 + 2 2 = 3 2. The following examle goes beyond the eighth grade standards, but is included to emhasize to students that the rules of arithmetic extend to exressions with root symbols in them. Examle 6. a. Solve: 3 x = 39. Solution. Dividing both sides by 3, we have the equation x = 13. Now squaring both sides, we get x = 13 2 = 169. 2014 University of Utah Middle School Math Project in artnershi with the 8MF7-6

b. Solve: 5 x + x = 24. Solution. 5 x + x = 6 x,, leading to the equation 6 x = 24,simlifying to x = 4. Squaring, we get the answer: x = 4 2 = 16. c. Which is greater: 14 or 10 2? Solution. We want to determine if the statement 14 < 10 2 is true or false. We could aroximate 2, but the truth of our answer deends uon the accuracy of our aroximation. A better strategy is to square both sides, since we know that squaring two ositive numbers reserves the relation between them. So our test becomes: is this true: 14 2 < 100(2). Since 14 2 = 196, the answer is: yes, this is true. d. Is there an integer between 4 3 and 5 2? Solution. To solve this, we need to square both sides and ask: is there a erfect square between these numbers. Now (4 3) 2 = 48 and (5 2) = 50, and the only integer between these two is 49. Since 49 = 7 2, the answer is yes: 7. Section 7.3: Rational and Irrational Numbers The Rational Number System Know that numbers that are not rational are called irrational. Understand informally that every number has a decimal exansion; for rational numbers, show that the decimal exansion reeats eventually, and convert a decimal exansion which reeats eventually into a rational number.8.ns.1. Know that 2 is irrational. 8.EE.2, second art. Use rational aroximations of irrational numbers to comare the size of irrational numbers, locate them aroximately on a number line diagram, and estimate the value of exressions (e.g., 2 ). For examle, by truncating the decimal exansion of 2, show that 2 is between 1 and 2, then between 1.4 and 1.5, and exlain how to continue on to get better aroximations. 8.NS.2 The relationshi of numbers and lengths and their reresentation as decimals was a significant art of seventh grade mathematics. We now summarize that discussion. The decimal reresentation of rational numbers is the natural extension of the base ten lace-value reresentation of whole numbers. A decimal is a sequence of digits with a dot (called the decimal oint) laced between a air of consecutive digits. The sequence of the digits to the left of the dot is the integral art of the number and those to the right comrise the fractional art. Each lace to the left of the dot reresents a ositive ower of ten; each lace to the right is a negative ower of ten. The digit tells us how many of that ower of ten are in the number. For examle: 231.785 = 2(10 2 ) + 3(10) + 1 + 7(10 1 + 8(10 2 ) + 5(10 3 ) = 2(100) + 3(10) + 1(1) + 7 10 + 8 100 + 5 1000 If there is no fractional art, this may be indicated by lacing a 0 after the decimal oint (as in 24.0). If there is no whole number art in a given numeral, a 0 is usually laced before the decimal oint (for examle 0.75). There are only finitely many digits before the decimal oint, but there may be infinitely many after the decimal oint. In Grade 7 students discussed the decimal reresentation of a fraction; found by long division. Thus we recall, as examles: 8MF7-7 2014 University of Utah Middle School Math Project in artnershi with the

3 4 = 75 100 = 7 10 + 5 100 = 0.75, 11 4 = 23 4 = 2 + 75 100 = 2 + 7 10 + 5 100 = 2.75, 2 = 0.66666666 = 0.6, 3 62 = 5.636363 = 5.63. 11 In the two, the overbar reresents the sequence of digits that are reeated over and over again, for as long as the student has the atience for long division. A decimal is said to be terminating if it ends after finitely many stes in the long division, as in the case of the first two examles above. It is said to be eventually reeating if, after some number of laces, tthere is a attern of digits that reeats over and over again, as in the last two examles. In Grade 7 students observed that the decimal exansion of a rational number always either terminates after a finite number of digits or eventually begins to reeat the same finite sequence of digits over and over. The converse is true: if a decimal terminates, or is eventually reeating, it describes a length corresonding to a rational number. We will now develo an understanding of why this is. First we start with terminating decimals. Examle 7. Convert 0.275 to a fraction. Solution. 0.275 = 2 10 + 7 100 + 5 1000 = 2 10 + 7 10 2 + 5 10 3. Now, if we ut these terms over a common denominator, we get 2(10 2 ) + 7(10) + 5 10 3 = 275 10 3 = 275 1000. In general, a terminating decimal is an integer lus a sum of fractions, all of whose denominators are owers of 10. By multilying each term by 10/10 as many times as necessary, we can ut all terms over the same denominator. We have shown that above and in the following. 0.67321 becomes and 67321 10 5 = 67231 100000, 0.0038 = 3 1000 + 8 10000 = 38 10000. We conclude that a terminating decimal leads to a fraction of the form A/10 e where A is an integer and e is a ositive integer. Notice that the exression A/10 e is not necessary in lowest terms: 275 10 = 5 5 11 3 2 2 2 5 5 5 = 11 2 2 2 5 = 11 40. 2014 University of Utah Middle School Math Project in artnershi with the 8MF7-8

More generally, start with any decimal: reresent it as a fraction with denominator a ower of 10. As we reduce to lowest terms we can only cancel 2 s and 5 s in the denominator, leaving only factors of 2 and 5. Examle 8. a. 25 = 5 2, so we should exect that 1/25 can be reresented by a terminating decimal. In fact, if we multily by 1 in its disguise as 2 2 2 2 we get: 1 25 = 4 100 = 0.04. b. Consider 1/200. Since 200 = 2 3 5 2, we lack a factor of 5 in the denominator to have a ower of 10. We fix this by multilying the fraction by 1 in the form 5 5 to get 5/1000 =.005. We conclude that a terminating decimal reresents a fraction whose denominator is a roduct of 2 s and 5 s. This is the converse of what we learned in Grade 7. Thus A terminating decimal leads to a fraction whose denominator is a roduct of 2 s and/or 5 s, and conversely, any such fraction is reresented by a terminating decimal. What about a fraction of the form /q, where q is not a roduct of 2 s and/or 5 s? In Grade 6 students learn that by long division (of by q) we can create a decimal exansion for /q to as many laces as we lease. In Grade 7 this is taken a bit further. Since each ste in the long division roduces a whole number remainder that is less than q, after at most q stes we must reeat a remainder already seen. From that oint on each digit of the long division reeats, and the rocess continues indefinitely in this way. For examle, 1/3 = 0.3333... for as long as we care to continue the division.. For 10 3 roduces a quotient of 3 with a remainder of 1, leading to a cyclical reeat of 10 3. In the way of review, let s look at a more comlicated examle. Examle 9. Find the decimal exansion of 157/660. Solution. Dividing 157 by 660 gives a quotient of 0.2 and a remainder of 25. Divide 25 by 660 to get a quotient of 0.03 and a remainder of 52. So far we have 157 5.2 = 0.23 + 660 660. 5.2 660 roduces a quotient of 0.007 and a remainder whose numerator is 58. Now division by 660 gives a quotient of.0008 and a remainder whose numerator is 52. Since we ve already seen that remainder, the long division will reeat itself and we ll again get a quotient of 8 and a remainder of 58. As this alternation of remainders of 52 and 58 continues forever, we can conclude that 157 = 0.23787878..., 660 with the sequence 78 reeating itself as often as we need. This is written as 157/660 = 0.2378, where the over line indicates continued reetition. Examle 10. Find the decimal exansion of 3/11. 8MF7-9 2014 University of Utah Middle School Math Project in artnershi with the

Solution. Ignoring decimal oints, the first division in the long division is 30 11, giving 2 with a remainder of 8. So, the second division is 80 11, giving 7 with a remainder of 3. Then the third division 30 11 gives a remainder of 8, which we ve seen before. We have entered the reeating cycle with 27 as the reeating number. Now take care of the decimal oint: 3/11 is less than 1 and bigger than 0.1, so the answer is 3/11 = 0.27 These examles give amle evidence that the long division rocedure will eventually go into a cycle. To summarize The decimal exansion of a fraction is terminating or eventually reeating - that is, after some initial sequence of digits, there a following set of digits (which may consist of zeroes) that reeats over and over. Exress Decimals as Fractions To comlete this set of ideas we now demonstrate the converse of the boxed exression above: a terminating or eventually reeating decimal reresents a fraction. We ve already observed that a terminating decimal is a fraction whose denominator is a ower of 10, so let s concentrate on reeating decimals. Examle 11. Start with: 0.33333..., or in short notation 0.3. Let a reresent this number. The e ect of multilying by 10 is to move the decimal to the right one digit. That is: 10a = 3.3 We now have the two equations: a = 0.3. 10a = 3.3. But recall the meaning of 3.3: it stands for 3 + 0.3 which is 3 + a. So the second equation is 10a = 3 + a. Solve for a to get a = 3/9 = 1/3. Here s a more comlicated examle: Examle 12. Convert 0.234 to a decimal. Solution. Set a = 0.234. Now multily by 1000, to get these two equations: a = 0.234. 1000a = 234.234. The second is the same as:1000a = 234 +.234; and becomes (after substitution) 1000a = 234 + a, from which we conclude: 999a = 234, so a = 234/999, which, in lowest terms, is 26/111. Alternatively, we could subtract the first equation from the third, getting to 999a = 234, and ultimately the same answer. 2014 University of Utah Middle School Math Project in artnershi with the 8MF7-10

Examle 13. Let s move on to decimals that are eventually reeating, such as 0.2654. First we take care of the reeating art: let b = 0.54, and follow the method of the receding examles to get the equation 100b = 54 + b, so b = 54/99 = 6/11 in lowest terms. Now: 0.2654 = 0.26 + 0.0054 = 26 100 + 1 26 (0.54) = 100 100 + 1 100 ( 6 11 ) = 1 100 (26 + 6 11 ) = 1 100 (292 11 ) = 73 275. For a final examle, let s convert 0.23. We have: 0.23 = 2 10 + 1 10! 1 = 7 3 30. Extension. Exand the Number System In the receding, for any oint on the line, we have discussed how to get a sequence of terminating decimals that rovide better and better estimates (in fact, each is within one-tenth of the given oint than its redecessor). If we think of a oint on the line as blur on the line (as it is in reality), then we can reeat this rocess until our decimal exansion is inside the blur, and then indistinguishable from the oint of interest. Better otical devices will shrink the blur, and so we ll have to aly our rocess a few more times. Now, in the idealization of the real world that is the realm of mathematics, this rocess may never end; and so we envision infinite decimal exansions that can be terminated at any time to give us the estimate that is as close as we can erceive with out most advanced otics. Thus the reeating decimals make sense: if we want a decimal aroximation of 1/3 that is correct within one-tenthousandth, we take 0.3333. If we want a decimal aroximation correct within 500 decimal oints, we take 0. followed by 500 3 s. Let us review the oerational descrition of the decimal exansion through measurement: Let a be a oint on the number line. Let N be the largest integer less than or equal to a; that is N ale a < N + 1. Now divide the integer between N and N + 1 into tenths, and let d 1 be the number of tenths between N and a. If this lands us right on a, then a = N + d 1 /10. If not, divide the interval between N + d 1 /10 and N + (d 1 + 1)/10 into hundredths and let d 2 be the number of hundredths below a, so that N + d 1 10 + d 2 100 ale a < N + d 1 10 + d 2 + 1 100. Do the same thing with thousandths and continue indefinitely - or at least as far as your measuring device can take you. Some decimal exansions are neither terminating, nor ultimately reeating, for examle 0.101001000100001000001..., where the number of 0 s between 1 s continues to increase by one each time, indefinitely. Another is obtained by writing down the sequence of ositive integers right after each other: 0.123456789101112131415.... Do such exressions really define (ideal) oints on the line? This question, in some form or another, has been around as long as numbers have been used to measure lengths. Today, we know that it is not a question that can 8MF7-11 2014 University of Utah Middle School Math Project in artnershi with the

be answered through common sense or logical deduction but, as an assertion that is fundamental to contemorary mathematics, must be acceted as a given (or, in some interretations, as art of the definition of the line). Such decimal exansions that are neither terminating or reeating cannot reresent a rational number, so are said to be irrational. We can continue to make u decimal exansions that are neither terminating nor reeating, and in that way illustrate more irrational numbers. But what is imortant to understand is that there are constructible lengths (like the sides of some tilted squares) that are irrational. End Extension The fact that we are heading for is this: for any whole number N, if N is not a erfect square, then N is not exressible as a fraction; that is, N is an irrational number. The existence of irrational numbers was discovered by the ancient Greeks (about 5th century BCE), and they were terribly uset by the discovery, since it was a basic tenet of theirs that numbers and length measures of line segments were di erent reresentations of the same idea. What haened is that a member of the Pythagorean society showed that it is imossible to exress the length of the side of the dashed square in Figure 2 ( 2) by a fraction. Here we ll try to describe the modern argument that actually roves more: If N is a whole number then either it is a erfect square or N is not a fraction; that is, N is irrational. Another way to say this is: For N a whole number, if N is a fraction, then it is a whole number. Another way of making this statement is this: If N is a whole number, and N is rational, then N is also a whole number. Extension. The Greeks knew how to show this for N = 2. Euclid gave a geometric roof (based on the concet of commensurable lengths, and Aristotle gave an algebraic roof, as follows: Suose that 2 is a rational number. Exress it as 2 = q in lowest terms (meaning that and q have no common integer factor). In articular, it cannot be the case that and q are both even. Now square both sides of the equation to get 2 = 2 q 2 so that 2q 2 = 2. This last equation tells us that 2 is even (it has 2 as a factor). But, since the square of an odd number is odd, that tells us that must be even: = 2r for some integer r. Then 2 = 4r 2, and utting that in the last equation above we get 2q 2 = 4r 2 so that q 2 = 2r 2, and q is thus also even. This contradicts the statement that 2 = /q in lowest terms, so we conclude that our original assumtion that 2 is equal to a ratio of integers was false. Therefore, since 2 can not be reresented as a ratio of integers, 2 is irrational. The argument for any whole number N (that it is either a erfect square or irrational) is similar, but uses more about the structure of whole numbers (in terms of rimes), It might be interesting to go into this, but it is beyond the 8th grade core. 2014 University of Utah Middle School Math Project in artnershi with the 8MF7-12

Examle 14. Show that 50 is irrational. Solution. 50 > 7 2 and 50 < 8 2, so 50 lies between 7 and 8, and thus cannot be an integer. Another way to see this is to factor 50 as 25 2, from which we conclude that 50 = 5 2, and 2 is irrational. Examle 15. is another irrational number that is defined geometrically. How do we give a numerical value to this number? The ancient Greek mathematician, Archimedes, was able to rovide a geometric way of aroximating the value of. Here we describe it briefly, using its definition as the quotient of the circumference of a circle by its diameter. In Figure 7, consider the triangles drawn to be continued around the circle 8 times. Now measure the lengths of the diameter (5.3 cm) and the lengths of the outer edge of the triangles shown triangles. The length of the inner one is 2 cm, and of the outer one is 2.3 cm. Now, reroducing the triangles in each half quadrant, we obtain two octagons, one with erimeter 8 2, which is less than the circumference of the circle and the other with erimeter 8 2.3 which is greater than the circumference of the circle. Since we know the formula C = D, where C is the circumference of a circle, and D its diameter, we obtain 8 2 < C = D = 5.3 and 8 2.3 > C = D = 5.3. After division by 5.3, these two inequalities give the estimate: 3.02 < <3.47. To increase the accuracy, we increase the number of sides of the aroximating olygon. Archimedes created an algorithm to calculate the olygon circumferences each time the number of sides is doubled. At the next ste (using a olygon with 16 sides), the rocedure roduces the estimate 22/7 for, which has an error less than 0.002. Figure 7 End Extension 8MF7-13 2014 University of Utah Middle School Math Project in artnershi with the

Aroximating the Value of Irrational Numbers Use rational aroximations of irrational numbers to comare the size of irrational numbers, locate them aroximately on a number line diagram, and estimate the value of exressions (e.g., 2 ). 8.NS.2 Since irrational numbers are reresented by decimals that are neither terminating nor reeating, we have to rely on the definition of the number to find fractions that are close to the irrational number - hoefully as close as we need the aroximation to be. So, as we saw above, Archimedes used the definition of to find a rational number (22/7) that is within 1/500 of. Even better, Archimedes described an algorithm to get estimates that are closer and closer - as far as we need them to be. But how about square roots? We know that 2 can be reresented as the diagonal of a right triangle with leg lengths equal to 1, so we can measure the length of that diagonal with a ruler. But the accuracy of that measure deends uon the detail in our ruler. We d rather have an arithmetic way to find square roots - or, to be more accurate, to aroximate them. For this, we return to the discussion of decimals in 7th grade. There, a method was described that found decimals that came as close as ossible to reresenting a given length. However, it too, deended uon our caacity to measure. What we want is a method to aroximate square roots, which, if reeated over and over again, gives us an estimate of the square root that comes as close to the exact length as we need it to be. Before calculators were available, a method for estimation of square roots was that of trial and error. With calculators, it is not so tedious - let us describe it. Examle 16. Aroximate 2 correct to three decimal laces. Solution. Since 1 2 = 1 and 2 2 = 4, we know that 2 is between 1 and 2. Let s try 1.5. 1.5 2 = 2.25, so 1.5 is too big. Now 1.4 2 = 1.96 so 1.4 is too small, but not by much: 1.4 is correct to one decimal lace. Let s try 1.41: 1.41 2 = 1.9881 and 1.42 2 = 2.0164. It looks like 1.42 2 is a little closer, so let s try 1.416: 1.416 2 = 2.005; so we try 1.415 2 = 2.002, still a little large, but quite close. To see that we are within three decimal laces, we check 1.414 2 = 1.9994. That is retty close, and less that 2. To check that 1.414 is correct to three decimal laces, we check half an additional oint uwards: 1.4145 2 = 2.0008, so the exact value of 2 is between 1.4140 and 1.4145, so 1.414 is correct to three decimal laces. Examle 17. Gregory wants to ut a fence around a square lot of land behind his house of 2000 sq.ft. How many linear feet of fence will he need? Solution. His daughter, a member of this class, tells him that the length of a side will be 2000, and since there are four sides, Gregory will need 4 2000 linear feet of fence. He objects that he can t go to the lumber yard asking for 4 2000 linear feet of fence; he needs a number. His daughter thinks, well that is a number, but understands that he needs a decimal aroximation She says: First we try to get close: The square of 30 is 900; so we try 40: it s square is 1600, and 50 2 = 2500, so our answer is between 40 and 50. Let s try 45; since 45 2 = 2025, we are close. Gregory says that for his urose, this is close enough, but he is curious and asks his daughter if she can do better. Sure, she says, and calculates 44 2 = 1936, and informs her father that the answer is between 44 and 45 - and maybe closer to 45. So she calculates the squares of number between 45.5 and 50: Number Square of number 44.5 1980.25 44.6 1989.16 44.7 1998.09 44.8 2007.04 2014 University of Utah Middle School Math Project in artnershi with the 8MF7-14

and concludes that each lot has to have side length between 44.7 feet and 44.8 feet, and much closer to 44.7. She suggests that he go with 44.72; that is he will need 4 44.72 = 178.88 linear feet of fence. Checking, she calculates that the square of 44.72 (accurate to two decimal oints) is 1999.88. Examle 18. Find the square root of 187 accurate to two decimal laces. Solution. Since 10 2 = 100, we try 11 2 = 121, 12 2 = 144, 13 2 = 169 and 14 2 = 196. From this we conclude that 187 is between 13 and 14, and robably a little closer to 14. Next, try 13.7: 13.7 2 = 187.69, a little too big. Now try 13.65: 13.65 2 = 187.005; almost there! Now, to be sure we are correct to 2 decimal oints, we calculate 13.64 2 = 186.05. Since 13.65 2 is so much closer to 187, we conclude that 13.65 is the answer, accurate to two decimal laces. Trial and error seems to work fairly well, so long as we start with a good guess, and have a calculator at hand. But, in the field, the engineer will want a way to do this with a calculator or comuter, where the comuter does the guessing. Such a method is called Newton s method, after Isaac Newton, one of the discoverers of the Calculus. Newton reasoned this way: suose that we have an estimate a for the square root of N, That is: a 2 N, where the symbol means is close to. Dividing by a, we have a N/a, so N/a is another estimate, about as good as a. He also noticed that a and N/a lie on oosite sides of N: as follows: suose a < N, so that a 2 < N. Multily both sides by N, giving Na 2 < N 2. Now divide both sides by a 2 to get: N < N2 a 2 or N < N a. Because of this, the average of a and N/a average should be an even better estimate. So he set a 0 = 1 2 a + N, a and then reeated the logic with a 0, and once again with the new estimate, until the oeration of taking a new average roduced the same answer, u to the desired number of decimal laces. The amazing fact is that Newton showed that this actually works and in a small number of stes, and can be alied to find solutions of a wide range of numeric roblems. Let s now illustrate inf the following examle. Examle 19. Find the square root of two accurate u to four decimal laces. Solution. We try 1 and 2, getting 1 2 = 1 and 2 2 = 4. Now, 2/1 = 2, so according to Newton, we should now try the average: 1.5. Since 2/(1.5) = 4/3, we next try the new average: a 0 = 1 2 1.5 + 4 3! = 17 12 = 1.416. This already is a retty good estimate. Extension. Now, let us bring in calculators or a comutational rogram like Excel so that we can reeat this rocess until the new estimate is the same as the second. Excel calculations for 2 (using Newton s method) are shown in the table of Figure 8. Our discussion brought us to the second line (1.416666667 is 17/6 correct to 9 decimal laces). The table continues with Newton s method, until the estimate (last column) stabilizes. Actually, it stabilized (to four decimal laces) in the third line, but we have continued the calculation, to show that there is 8MF7-15 2014 University of Utah Middle School Math Project in artnershi with the

no further change. In fact each ste in Newton s method roduces a more accurate estimate, so once we have no change (to the number of correct decimals we require), there is no need to go further. First estimate: a = 1 Estimate 2/a New Estimate (Average) 1 2 1.5 1.5 1.333333 1.416666667 1.41667 1.411765 1.414215686 1.41422 1.414211 1.414213562 1.41421 1.414214 1.414213562 1.41421 1.414214 1.414213562 1.41421 1.414214 1.414213562 1.41421 1.414214 1.414213562 Figure 8: Square Root of 2 Examle 20. Use Newton s method to find the square root of 5. Solution. First of all find the square root of five is not very meaningful. Using the tilted square with side lengths 1 and 2, we found the square root of 5 as a length. So, erhas here wen mean, find the numerical value of the square root of 5. But as we have already observed, 5 is not exressible as a fraction, so we can t exect to find its value recisely. What we can hoe for is to find a decimal exansion that comes as close as we lease to 5. So, let us make the question recise: find a decimal that is an estimate of 5 that is correct to 4 decimal laces. Let s go through Newton s method. First we see that 2 is a good aroximation for 5 by an integer, since 2 2 = 4 and 3 2 = 9.Now 5/2 = 2.5 is also a good guess, since 2.5 2 = 6.25. In fact it is what Newton s method wants us to look at next. The rest of the comutation was done by Excel and is shown in Figure 8a. There we see that by the third ste in Newton s method, we have 5 correct to 6 decimal laces. First estimate: a = 2 Estimate 5/a New Estimate (Average) 2 2.5 2.25 2.25 2.222222 2.236111111 2.23611 2.236025 2.236067978 2.23607 2.236068 2.236067977 2.23607 2.236068 2.236067977 2.23607 2.236068 2.236067977 2.23607 2.236068 2.236067977 2.23607 2.236068 2.236067977 Figure 9: Square Root of 5 Do we ever get to the lace where the last two aroximations agree erfectly - that is: when have we arrived at the numerical value of the desired oint? Alas, for irrational numbers, the answer is we never do, but we do get better and better. Unless N itself is a erfect square (the square of another integer), we will never arrive at a decimal exansion that is recisely N. But - and this is all we really need - we can get as close as we want. Examle 21. Now use Newton s method to calculate 25 and 1000. Solution. The calculation is exhibited in Figure 10a and 10b resectively. 2014 University of Utah Middle School Math Project in artnershi with the 8MF7-16

First estimate: a = 1 Estimate 25/a New Estimate (Average) 1 25 13 13 1.923077 7.461538462 7.46154 3.350515 5.406026963 5.40603 4.624468 5.015247602 5.01525 4.984799 5.000023178 5.00002 4.999977 5 5 5 5 Figure 10a: Square Root of 25 First estimate: a = 25 Estimate 1000/a New Estimate (Average) 25 40 32.5 32.5 30.76923 31.63461538 31.6346 31.61094 31.62277882 31.6228 31.62277 31.6227766 31.6228 31.62278 31.6227766 31.6228 31.62278 31.6227766 31.6228 31.62278 31.6227766 31.6228 31.62278 31.6227766 Figure 10b: Square Root of 1000 End Extension. Examle 22. Find a simler algebraic exression for 1000. Solution. To be clear, we are asked to factor 1000 in such a way so as to ull out as large a erfect square as ossible. Since 1000 = 100 10, and 100 is a erfect square, the answer is 1000 = 10 10 = 10 2 5. We could argue that, to aroximate 1000, instead of alying a square root method to 1000, we could use the aroximations of 2 and 5 that we already have and multily. The aroximations we already have (see Figures 8 and 9) are: 2 = 1.414213562 and 5 = 2.236067977. Multilying, we find: 1000 = 10 2 5 = (10)(1.414213562)(2.236067977) = 31.6227752. Now, comare this with Figure 10b and note the loss of accuracy in the sixth and seventh decimal laces. This is not an error in multilication (which was done on a basic hand calculator), but a fact about multilying decimal aroximations: always exect a loss of accuracy in the last decimal laces. The closeness of aroximations changes as we add or multily them. So, if A 0 is a 2-decimal aroximation to A and B 0 ais the same for B, we cannot conclude that A 0 + B 0 is a 2-decimal aroximation to A + B, nor that A 0 B is a 2-decimal aroximation to AB. Let us illustrate that. 8MF7-17 2014 University of Utah Middle School Math Project in artnershi with the

Examle 23. 3.16 agrees with 10 to two decimal laces. But 3.16 3.16 = 9.9856, which does not agree with 10 to two decimal laces. Examle 24. Aroximate + 2 to three decimal laces. Start with the aroximations u to four decimal oints: 3.1416 and 1.4142, and add: 3.1416+1.4142 = 4.5558, from which we accet 4.556 as the three (not four) decimal aroximation because of the rounding to the fourth lace in the original aroximations. For roducts it is not so easy. Suose that A 0 and B 0 are aroximations to two articular numbers, which we can denote as A, B, Then we have A = A 0 + e 1, B = B 0 + e 0 where e and e 0 are the errors in aroximation. For the roduct we will have AB = (A 0 + e)(b 0 + e 0 ) = A 0 B 0 + A 0 e 0 + B 0 e + ee 0, telling us that the magnitude of the error has been multilied by the factors A and B, which may ut us very far from the desired degree of accuracy. Examle 25. Suose that A and B aroximate the numbers 1 and 100 resectively within one decimal oint. That means that 0.95 < A < 1.05 and 99.95 < B < 100.05. When we multily, we find that AB could anywhere in the region 94.9525 < AB < 105.0525, thus, as much as 5 units away from the accurate roduct 100. To find aroximate values for irrational numbers, we have to understand the definition of the number so we can use it for this urose. For examle, suose we want to find a number whose cube is 35, correct u to two decimal laces. Start with a good guess. Since 3 3 = 27 and 4 3 = 64, we take 3 as our first guess. Since 27 is much closer to 35 than 64, we now calculate the cubes of 3.1, 3.2, 3.3,... until we find those closest to 35: 3.1 3 = 29.791, 3.2 3 = 32.6768, 3.3 3 = 35.937. We can sto here, since the last calculated number is larger than 35. Now we go to the next decimal lace, starting at 3.3, working down (since 35.937 is closer to 35 than 32.6768: 3.29 3 = 35.611, 3.28 3 = 35.287, 3.27 3 = 34.965. Since the last number is as close as we can get to 35 with two lace decimals, we conclude that 3.27 is correct to two decimal laces. Examle 26. What is bigger 2 or 10? Solution. In the days before comuters, the high school hysics teacher recommended to students that they could use 10 as an aoroximation to 2 so as to simlify some calculations when neceesary, Here we ask: which is bigger? Today we use 22/7 as an aroximation to, so let s try (22/7) 2. That is equal to 9.877 which is su ciently less than 10, so we can susect that 2 < 10. Now, recall from examle 3 that 3.16 2 = 9.9856. Since we know that 3.16 > f rac227 = 3.1428..., we have stronger evidence that 2 < 10. In fact, a seven-decimal aroximation to is 3.141529, so we can conclude that < 22 7 < 3.16 and thus 2 < 10 2014 University of Utah Middle School Math Project in artnershi with the 8MF7-18